3.171 \(\int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=107 \[ \frac {\sin ^5(c+d x) \sqrt {\cos (c+d x)}}{5 d \sqrt {b \cos (c+d x)}}-\frac {2 \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}} \]

[Out]

sin(d*x+c)*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)-2/3*sin(d*x+c)^3*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+1/
5*sin(d*x+c)^5*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 2633} \[ \frac {\sin ^5(c+d x) \sqrt {\cos (c+d x)}}{5 d \sqrt {b \cos (c+d x)}}-\frac {2 \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(11/2)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]]) - (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Sqrt[b
*Cos[c + d*x]]) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[b*Cos[c + d*x]])

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \cos ^5(c+d x) \, dx}{\sqrt {b \cos (c+d x)}}\\ &=-\frac {\sqrt {\cos (c+d x)} \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {b \cos (c+d x)}}\\ &=\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 57, normalized size = 0.53 \[ \frac {\sin (c+d x) \left (3 \sin ^4(c+d x)-10 \sin ^2(c+d x)+15\right ) \sqrt {\cos (c+d x)}}{15 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(11/2)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(Sqrt[Cos[c + d*x]]*Sin[c + d*x]*(15 - 10*Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15*d*Sqrt[b*Cos[c + d*x]])

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fricas [A]  time = 0.67, size = 54, normalized size = 0.50 \[ \frac {{\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, b d \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*cos(d*x + c)^4 + 4*cos(d*x + c)^2 + 8)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(b*d*sqrt(cos(d*x + c)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {11}{2}}}{\sqrt {b \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(11/2)/sqrt(b*cos(d*x + c)), x)

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maple [A]  time = 0.12, size = 52, normalized size = 0.49 \[ \frac {\left (3 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )+8\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{15 d \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x)

[Out]

1/15/d*(3*cos(d*x+c)^4+4*cos(d*x+c)^2+8)*sin(d*x+c)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)

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maxima [A]  time = 1.43, size = 68, normalized size = 0.64 \[ \frac {3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )}{240 \, \sqrt {b} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/240*(3*sin(5*d*x + 5*c) + 25*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*sin(1/5*arctan2(sin(
5*d*x + 5*c), cos(5*d*x + 5*c))))/(sqrt(b)*d)

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mupad [B]  time = 1.18, size = 73, normalized size = 0.68 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (175\,\sin \left (2\,c+2\,d\,x\right )+28\,\sin \left (4\,c+4\,d\,x\right )+3\,\sin \left (6\,c+6\,d\,x\right )\right )}{240\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(11/2)/(b*cos(c + d*x))^(1/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(175*sin(2*c + 2*d*x) + 28*sin(4*c + 4*d*x) + 3*sin(6*c + 6*d*x)))/
(240*b*d*(cos(2*c + 2*d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(11/2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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